Exponential distribution

The exponential distribution models the waiting time until the next event when events occur at a constant rate.

f(x) = \lambda e^{-\lambda x}, \qquad P(X > x) = e^{-\lambda x}, \qquad x \ge 0

pdf and survival function

$\lambda$: rate: average number of events per unit of time
$x$: waiting time

\E(X) = \frac{1}{\lambda}, \qquad \SD(X) = \frac{1}{\lambda}, \qquad \text{median} = \frac{\ln 2}{\lambda}

Mean, SD and median

$1/\lambda$: mean waiting time

When to use

Times between events of a Poisson process: time to the next call, time between failures of a component with no wear-out.

Assumptions

A constant event rate. If the rate changes over time (wear-out, rush hours), the exponential is the wrong model.

How to interpret

Memoryless: $P(X > s + t \given X > s) = P(X > t)$ . Having already waited does not shorten the remaining wait.
The median is below the mean (right skew): most waits are shorter than average.
Link to Poisson: if counts per unit time are Poisson( $\lambda$ ), gaps are exponential with rate $\lambda$ .

Common mistakes

Parameterising by the mean when software expects the rate (or vice versa). The engine uses the rate.
Expecting a long wait to make an event "overdue".

Example

Calls arrive at a rate of $\lambda = 0.25$ per minute (mean gap 4 minutes). $P(X > 6) = e^{-1.5} = 0.2231$ ; $P(X \le 2) = 0.3935$ ; median wait $= 2.7726$ minutes. Memorylessness: $P(X > 10 \given X > 4) = P(X > 6) = 0.2231$ .